Question

Bleaching powder and bleach solution are produced on a large scale and used in several household products. The effectiveness of bleach solution is often measured by iodometry.
$25 \,mL$ of household bleach solution was mixed with $30 \,mL$ of $0.50 \,M\, Kl$ and $10 \,mL$ of $4 \,N$ acetic acid. In the titration of the liberated iodine, $48\, mL$ of $0.25\, N Na _{2} S _{2} O _{3}$ was used to reach the end point. The molarity of the household bleach solution is

• A. 0.48 M
• B. 0.96 M
• C. 0.24 M
• D. 0.024 M

Answer 1

Answer: (C)

Consider the titration reaction
$CaOCl _{2}+2 KI \rightarrow I _{2}+ Ca ( OH )_{2}+ KCl$
According to this reaction, $25 \,mL$ of $CaOCl _{2}$, reacts with $30\, mL$ of $0.50 \,M\,KI$
$I _{2}+2 Na _{2} S _{2} O _{3} \rightarrow Na _{2} S _{4} O _{6}+2 NaI$
Given that $48\, mL$ of $0.25 \,N\, Na _{2} S _{2} O _{3}$ was used to reach the end point.
So, the number of moles of $I _{2}$ produced $=48 \times 0.25 / 2=6$.
According to the reaction,
Number of millimoles of bleaching powder $=$ Number of moles of $I_{2}$
$= (1 / 2) \times$ Number of moles of $Na _{2} S _{2} O _{3}=6$
So, the molarity of $CaOCl _{2}$ is
$\frac{\text { Number of moles of bleaching powder }}{\text { Volume of solution }}$
$=6 \frac{\text { m\,mol }}{25\, mL }=0.24 \,M$